﻿ POJ – 2406 Power Strings（KMP） | 玄奇博客 # POJ – 2406 Power Strings（KMP）

### Power Strings

Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 58869 Accepted: 24457

### Description

Given two strings a and b we define ab to be their concatenation. For example, if a = “abc” and b = “def” then ab = “abcdef”. If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = “” (the empty string) and a^(n+1) = a*(a^n).

### Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

### Output

For each s you should print the largest n such that s = a^n for some string a.

### Sample Input

abcd
aaaa
ababab
.


### Sample Output

1
4
3


### Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

### Source

Waterloo local 2002.07.01

### 代码

#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>

using namespace std;

const int maxn = 1001000;

char arrn[maxn];

void preKMP(char x[], int m, int kmpNext[]){
int i, j;
j=kmpNext=-1;
i=0;
while(i < m){
while(-1 != j && x[i] != x[j]) j = kmpNext[j];
kmpNext[++i] = ++j;
}
}

int nxt[maxn];

int main(){
//freopen("in.txt", "r", stdin);
int n;
while(~scanf("%s", arrn)){
if(arrn == '.') break;
int n = strlen(arrn);
preKMP(arrn, n, nxt);
int len = n - nxt[n];
if( n % len )printf("1\n");  //不能表示为循环节
else printf("%d\n",  n/len);
}

return 0;
}